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Ulleksa [173]
4 years ago
10

Brainliest to correct answer

Mathematics
1 answer:
Gnesinka [82]4 years ago
4 0

Question 1:

For this case we must factor the following expression:

n ^ 2 + 2n-24 = 0

We must find two numbers that when multiplied result in -24 and when added together result in 2. These numbers are: 6 and -4.

6*(-4)=-24\\6-4=2

Thus, we factor:

(x + 6) (x-4) = 0

The roots are:

x_ {1} = - 6\\x_ {2} = 4

Answer:

Option C

Question 2:

For this case we have that by definition, the area of a rectangle is given by:

A = w * l

Where:

w: Is the width of the rectangle

l: is the length of the rectangle

According to the statement we have:

A = 30 \ ft ^ 2\\l = 3w-1

Substituting:

(3w-1) w = 30\\3w ^ 2-w = 30\\3w ^ 2-w-30 = 0

Where:

a = 3\\b = -1\\c = -30

The solution:

x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}\\x = \frac {- (- 1) \pm \sqrt {(- 1) ^ 2-4 (3) (- 30)}} {2 (3)}\\x = \frac {1 \pm \sqrt {1 + 360}} {2 (3)}\\x = \frac {1 \pm \sqrt {361}} {6}\\x = \frac {1\pm19} {6}\\

We have two roots:

x_ {1} = \frac {1-19} {6} = \frac {-18} {6} = - 3\\x_ {2} = \frac {1 + 19} {6} =\frac {20} {6} = 3,333

We choose the second value (the positive), when rounding is 3.33 \ ft

Answer:

Option A

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Read 2 more answers
Simplify each rational expression to lowest terms, specifying the values of xx that must be excluded to avoid division
k0ka [10]

Answer:

(a) \frac{x^2-6x+5}{x^2-3x-10}=\frac{x-1}{x+2}. The domain of this function is all real numbers not equal to -2 or 5.

(b) \frac{x^3+3x^2+3x+1}{x^3+2x^2-x}=1+\frac{x^2+4x+1}{x^3+2x^2-x}. The domain of this function is all real numbers not equal to 0, -1+\sqrt{2} or -1+\sqrt{2}.

(c) \frac{x^2-16}{x^2+2x-8}=\frac{x-4}{x-2}.The domain of this function is all real numbers not equal to 2 or -4.

(d) \frac{x^2-3x-10}{x^3+6x^2+12x+8}=\frac{x-5}{\left(x+2\right)^2}. The domain of this function is all real numbers not equal to -2.

(e) \frac{x^3+1}{x^2+1}=x+\frac{-x+1}{x^2+1}. The domain of this function is all real numbers.

Step-by-step explanation:

To reduce each rational expression to lowest terms you must:

(a) For \frac{x^2-6x+5}{x^2-3x-10}

\mathrm{Factor}\:x^2-6x+5\\\\x^2-6x+5=\left(x^2-x\right)+\left(-5x+5\right)\\x^2-6x+5=x\left(x-1\right)-5\left(x-1\right)\\\\\mathrm{Factor\:out\:common\:term\:}x-1\\x^2-6x+5=\left(x-1\right)\left(x-5\right)

\mathrm{Factor}\:x^2-3x-10\\\\x^2-3x-10=\left(x^2+2x\right)+\left(-5x-10\right)\\x^2-3x-10=x\left(x+2\right)-5\left(x+2\right)\\\\\mathrm{Factor\:out\:common\:term\:}x+2\\x^2-3x-10=\left(x+2\right)\left(x-5\right)

\frac{x^2-6x+5}{x^2-3x-10}=\frac{\left(x-1\right)\left(x-5\right)}{\left(x+2\right)\left(x-5\right)}

\mathrm{Cancel\:the\:common\:factor:}\:x-5\\\\\frac{x^2-6x+5}{x^2-3x-10}=\frac{x-1}{x+2}

The denominator in a fraction cannot be zero because division by zero is undefined. So we need to figure out what values of the variable(s) in the expression would make the denominator equal zero.

To find any values for x that would make the denominator = 0 you need to set the denominator = 0 and solving the equation.

x^2-3x-10=\left(x+2\right)\left(x-5\right)=0

Using the Zero Factor Theorem: = 0 if and only if = 0 or = 0

x+2=0\\x=-2\\\\x-5=0\\x=5

The domain is the set of all possible inputs of a function which allow the function to work. Therefore the domain of this function is all real numbers not equal to -2 or 5.

(b) For \frac{x^3+3x^2+3x+1}{x^3+2x^2-x}

\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}x^3+3x^2+3x+1\mathrm{\:and\:the\:divisor\:}x^3+2x^2-x\mathrm{\::\:}\frac{x^3}{x^3}=1

Quotient = 1

\mathrm{Multiply\:}x^3+2x^2-x\mathrm{\:by\:}1:\:x^3+2x^2-x

\mathrm{Subtract\:}x^3+2x^2-x\mathrm{\:from\:}x^3+3x^2+3x+1\mathrm{\:to\:get\:new\:remainder}

Remainder = x^2+4x+1}

\frac{x^3+3x^2+3x+1}{x^3+2x^2-x}=1+\frac{x^2+4x+1}{x^3+2x^2-x}

  • The domain of this function is all real numbers not equal to 0, -1+\sqrt{2} or -1+\sqrt{2}.

x^3+2x^2-x=0\\\\x^3+2x^2-x=x\left(x^2+2x-1\right)=0\\\\\mathrm{Solve\:}\:x^2+2x-1=0:\quad x=-1+\sqrt{2},\:x=-1-\sqrt{2}

(c) For \frac{x^2-16}{x^2+2x-8}

x^2-16=\left(x+4\right)\left(x-4\right)

x^2+2x-8= \left(x-2\right)\left(x+4\right)

\frac{x^2-16}{x^2+2x-8}=\frac{\left(x+4\right)\left(x-4\right)}{\left(x-2\right)\left(x+4\right)}\\\\\frac{x^2-16}{x^2+2x-8}=\frac{x-4}{x-2}

  • The domain of this function is all real numbers not equal to 2 or -4.

x^2+2x-8=0\\\\x^2+2x-8=\left(x-2\right)\left(x+4\right)=0

(d) For \frac{x^2-3x-10}{x^3+6x^2+12x+8}

\mathrm{Factor}\:x^2-3x-10\\\left(x^2+2x\right)+\left(-5x-10\right)\\x\left(x+2\right)-5\left(x+2\right)

\mathrm{Apply\:cube\:of\:sum\:rule:\:}a^3+3a^2b+3ab^2+b^3=\left(a+b\right)^3\\\\a=x,\:\:b=2\\\\x^3+6x^2+12x+8=\left(x+2\right)^3

\frac{x^2-3x-10}{x^3+6x^2+12x+8}=\frac{\left(x+2\right)\left(x-5\right)}{\left(x+2\right)^3}\\\\\frac{x^2-3x-10}{x^3+6x^2+12x+8}=\frac{x-5}{\left(x+2\right)^2}

  • The domain of this function is all real numbers not equal to -2

x^3+6x^2+12x+8=0\\\\x^3+6x^2+12x+8=\left(x+2\right)^3=0\\x=-2

(e) For \frac{x^3+1}{x^2+1}

\frac{x^3+1}{x^2+1}=x+\frac{-x+1}{x^2+1}

  • The domain of this function is all real numbers.

x^2+1=0\\x^2=-1\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{-1},\:x=-\sqrt{-1}

4 0
3 years ago
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